141373
The displacement \(x\) of a particle varies with time \(t\) as \(x=a e^{-\alpha t}+b e^{\beta t}\), where \(a, b, \alpha\) and \(\beta\) are positive constants. The velocity of the particle will
1 go on decreasing with time
2 be independent of \(\alpha\) and \(\beta\)
3 drop to zero when \(\alpha=\beta\)
4 go on increasing with time
Explanation:
D Given, \(x=a e^{-\alpha t}+b e^{\beta t}\) Differentiating w.r.t. \(t\), we get \(\frac{d x}{d t}=a e^{-\alpha t}(-\alpha)+b e^{\beta t} \cdot \beta\) \(v=-a \alpha e^{-\alpha t}+b \beta e^{\beta t}\) From the above equation we can tell that velocity will increase with time since the negative component will decrease with an increase in time and the positive component will increase in the given equation.
JIPMER- 2007
Motion in One Dimensions
141190
An object is released from a height of \(h\) and after rebound it attains a height of \(\frac{h}{2}\). Which of the following velocity (V) vs. height (H) graphs describes this journey correctly? (velocity in upward direction is positive)
1
2
3
4
Explanation:
A \(\mathrm{V}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) downward velocity is negative and upward is positive when ball is coming down graph will be in IV quadrant i.e. \(\mathrm{v}\) is -ve and when going up graph will be in I quadrant i.e. \(\mathrm{v}\) is \(+\mathrm{ve}\).
Tripura-2021
Motion in One Dimensions
141211
The graph which represents the velocity time dependence of a solid descending in a viscous sodium is
1
2
3
4
Explanation:
B Initially the velocity of the body falling in a viscous fluid increases but after sometimes, the body moves with a constant velocity known as terminal velocity.
Tripura-2020
Motion in One Dimensions
141256
An athlete runs exactly once around a circular track of length \(500 \mathrm{~m}\). The runner's displacement in the race is
1 \(50 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(0.5 \mathrm{~m}\)
4 \(0 \mathrm{~m}\)
Explanation:
D Displacement is known as the distance between initial position and final position. In a circular track, the initial and final position of the athlete is same. The total distance covered is \(500 \mathrm{~m}\), but the net displacement is zero. So, total displacement is zero.
J and K-CET-2018
Motion in One Dimensions
141270
The graph which cannot possibly represent one-dimensional motion is
1
2
3
4
5
Explanation:
C In one dimension motion, At \(\mathrm{t}=0\), the body must at one position in graph. Hence option c has two positions in the graph at \(\mathrm{t}=0\), it will be correct option.
141373
The displacement \(x\) of a particle varies with time \(t\) as \(x=a e^{-\alpha t}+b e^{\beta t}\), where \(a, b, \alpha\) and \(\beta\) are positive constants. The velocity of the particle will
1 go on decreasing with time
2 be independent of \(\alpha\) and \(\beta\)
3 drop to zero when \(\alpha=\beta\)
4 go on increasing with time
Explanation:
D Given, \(x=a e^{-\alpha t}+b e^{\beta t}\) Differentiating w.r.t. \(t\), we get \(\frac{d x}{d t}=a e^{-\alpha t}(-\alpha)+b e^{\beta t} \cdot \beta\) \(v=-a \alpha e^{-\alpha t}+b \beta e^{\beta t}\) From the above equation we can tell that velocity will increase with time since the negative component will decrease with an increase in time and the positive component will increase in the given equation.
JIPMER- 2007
Motion in One Dimensions
141190
An object is released from a height of \(h\) and after rebound it attains a height of \(\frac{h}{2}\). Which of the following velocity (V) vs. height (H) graphs describes this journey correctly? (velocity in upward direction is positive)
1
2
3
4
Explanation:
A \(\mathrm{V}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) downward velocity is negative and upward is positive when ball is coming down graph will be in IV quadrant i.e. \(\mathrm{v}\) is -ve and when going up graph will be in I quadrant i.e. \(\mathrm{v}\) is \(+\mathrm{ve}\).
Tripura-2021
Motion in One Dimensions
141211
The graph which represents the velocity time dependence of a solid descending in a viscous sodium is
1
2
3
4
Explanation:
B Initially the velocity of the body falling in a viscous fluid increases but after sometimes, the body moves with a constant velocity known as terminal velocity.
Tripura-2020
Motion in One Dimensions
141256
An athlete runs exactly once around a circular track of length \(500 \mathrm{~m}\). The runner's displacement in the race is
1 \(50 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(0.5 \mathrm{~m}\)
4 \(0 \mathrm{~m}\)
Explanation:
D Displacement is known as the distance between initial position and final position. In a circular track, the initial and final position of the athlete is same. The total distance covered is \(500 \mathrm{~m}\), but the net displacement is zero. So, total displacement is zero.
J and K-CET-2018
Motion in One Dimensions
141270
The graph which cannot possibly represent one-dimensional motion is
1
2
3
4
5
Explanation:
C In one dimension motion, At \(\mathrm{t}=0\), the body must at one position in graph. Hence option c has two positions in the graph at \(\mathrm{t}=0\), it will be correct option.
141373
The displacement \(x\) of a particle varies with time \(t\) as \(x=a e^{-\alpha t}+b e^{\beta t}\), where \(a, b, \alpha\) and \(\beta\) are positive constants. The velocity of the particle will
1 go on decreasing with time
2 be independent of \(\alpha\) and \(\beta\)
3 drop to zero when \(\alpha=\beta\)
4 go on increasing with time
Explanation:
D Given, \(x=a e^{-\alpha t}+b e^{\beta t}\) Differentiating w.r.t. \(t\), we get \(\frac{d x}{d t}=a e^{-\alpha t}(-\alpha)+b e^{\beta t} \cdot \beta\) \(v=-a \alpha e^{-\alpha t}+b \beta e^{\beta t}\) From the above equation we can tell that velocity will increase with time since the negative component will decrease with an increase in time and the positive component will increase in the given equation.
JIPMER- 2007
Motion in One Dimensions
141190
An object is released from a height of \(h\) and after rebound it attains a height of \(\frac{h}{2}\). Which of the following velocity (V) vs. height (H) graphs describes this journey correctly? (velocity in upward direction is positive)
1
2
3
4
Explanation:
A \(\mathrm{V}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) downward velocity is negative and upward is positive when ball is coming down graph will be in IV quadrant i.e. \(\mathrm{v}\) is -ve and when going up graph will be in I quadrant i.e. \(\mathrm{v}\) is \(+\mathrm{ve}\).
Tripura-2021
Motion in One Dimensions
141211
The graph which represents the velocity time dependence of a solid descending in a viscous sodium is
1
2
3
4
Explanation:
B Initially the velocity of the body falling in a viscous fluid increases but after sometimes, the body moves with a constant velocity known as terminal velocity.
Tripura-2020
Motion in One Dimensions
141256
An athlete runs exactly once around a circular track of length \(500 \mathrm{~m}\). The runner's displacement in the race is
1 \(50 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(0.5 \mathrm{~m}\)
4 \(0 \mathrm{~m}\)
Explanation:
D Displacement is known as the distance between initial position and final position. In a circular track, the initial and final position of the athlete is same. The total distance covered is \(500 \mathrm{~m}\), but the net displacement is zero. So, total displacement is zero.
J and K-CET-2018
Motion in One Dimensions
141270
The graph which cannot possibly represent one-dimensional motion is
1
2
3
4
5
Explanation:
C In one dimension motion, At \(\mathrm{t}=0\), the body must at one position in graph. Hence option c has two positions in the graph at \(\mathrm{t}=0\), it will be correct option.
141373
The displacement \(x\) of a particle varies with time \(t\) as \(x=a e^{-\alpha t}+b e^{\beta t}\), where \(a, b, \alpha\) and \(\beta\) are positive constants. The velocity of the particle will
1 go on decreasing with time
2 be independent of \(\alpha\) and \(\beta\)
3 drop to zero when \(\alpha=\beta\)
4 go on increasing with time
Explanation:
D Given, \(x=a e^{-\alpha t}+b e^{\beta t}\) Differentiating w.r.t. \(t\), we get \(\frac{d x}{d t}=a e^{-\alpha t}(-\alpha)+b e^{\beta t} \cdot \beta\) \(v=-a \alpha e^{-\alpha t}+b \beta e^{\beta t}\) From the above equation we can tell that velocity will increase with time since the negative component will decrease with an increase in time and the positive component will increase in the given equation.
JIPMER- 2007
Motion in One Dimensions
141190
An object is released from a height of \(h\) and after rebound it attains a height of \(\frac{h}{2}\). Which of the following velocity (V) vs. height (H) graphs describes this journey correctly? (velocity in upward direction is positive)
1
2
3
4
Explanation:
A \(\mathrm{V}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) downward velocity is negative and upward is positive when ball is coming down graph will be in IV quadrant i.e. \(\mathrm{v}\) is -ve and when going up graph will be in I quadrant i.e. \(\mathrm{v}\) is \(+\mathrm{ve}\).
Tripura-2021
Motion in One Dimensions
141211
The graph which represents the velocity time dependence of a solid descending in a viscous sodium is
1
2
3
4
Explanation:
B Initially the velocity of the body falling in a viscous fluid increases but after sometimes, the body moves with a constant velocity known as terminal velocity.
Tripura-2020
Motion in One Dimensions
141256
An athlete runs exactly once around a circular track of length \(500 \mathrm{~m}\). The runner's displacement in the race is
1 \(50 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(0.5 \mathrm{~m}\)
4 \(0 \mathrm{~m}\)
Explanation:
D Displacement is known as the distance between initial position and final position. In a circular track, the initial and final position of the athlete is same. The total distance covered is \(500 \mathrm{~m}\), but the net displacement is zero. So, total displacement is zero.
J and K-CET-2018
Motion in One Dimensions
141270
The graph which cannot possibly represent one-dimensional motion is
1
2
3
4
5
Explanation:
C In one dimension motion, At \(\mathrm{t}=0\), the body must at one position in graph. Hence option c has two positions in the graph at \(\mathrm{t}=0\), it will be correct option.
141373
The displacement \(x\) of a particle varies with time \(t\) as \(x=a e^{-\alpha t}+b e^{\beta t}\), where \(a, b, \alpha\) and \(\beta\) are positive constants. The velocity of the particle will
1 go on decreasing with time
2 be independent of \(\alpha\) and \(\beta\)
3 drop to zero when \(\alpha=\beta\)
4 go on increasing with time
Explanation:
D Given, \(x=a e^{-\alpha t}+b e^{\beta t}\) Differentiating w.r.t. \(t\), we get \(\frac{d x}{d t}=a e^{-\alpha t}(-\alpha)+b e^{\beta t} \cdot \beta\) \(v=-a \alpha e^{-\alpha t}+b \beta e^{\beta t}\) From the above equation we can tell that velocity will increase with time since the negative component will decrease with an increase in time and the positive component will increase in the given equation.
JIPMER- 2007
Motion in One Dimensions
141190
An object is released from a height of \(h\) and after rebound it attains a height of \(\frac{h}{2}\). Which of the following velocity (V) vs. height (H) graphs describes this journey correctly? (velocity in upward direction is positive)
1
2
3
4
Explanation:
A \(\mathrm{V}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) downward velocity is negative and upward is positive when ball is coming down graph will be in IV quadrant i.e. \(\mathrm{v}\) is -ve and when going up graph will be in I quadrant i.e. \(\mathrm{v}\) is \(+\mathrm{ve}\).
Tripura-2021
Motion in One Dimensions
141211
The graph which represents the velocity time dependence of a solid descending in a viscous sodium is
1
2
3
4
Explanation:
B Initially the velocity of the body falling in a viscous fluid increases but after sometimes, the body moves with a constant velocity known as terminal velocity.
Tripura-2020
Motion in One Dimensions
141256
An athlete runs exactly once around a circular track of length \(500 \mathrm{~m}\). The runner's displacement in the race is
1 \(50 \mathrm{~m}\)
2 \(5 \mathrm{~m}\)
3 \(0.5 \mathrm{~m}\)
4 \(0 \mathrm{~m}\)
Explanation:
D Displacement is known as the distance between initial position and final position. In a circular track, the initial and final position of the athlete is same. The total distance covered is \(500 \mathrm{~m}\), but the net displacement is zero. So, total displacement is zero.
J and K-CET-2018
Motion in One Dimensions
141270
The graph which cannot possibly represent one-dimensional motion is
1
2
3
4
5
Explanation:
C In one dimension motion, At \(\mathrm{t}=0\), the body must at one position in graph. Hence option c has two positions in the graph at \(\mathrm{t}=0\), it will be correct option.
